Question

Find all numbers c that satisfy the conclusion of the Mean Value Theorem for the following function and interval. Enter the values in increasing order and enter N in any blanks you don't need to use.

f(x)= e^{-7 x}, [0,9]

Answer 1

Anna:
Have you considered the conditions that must be satisfied before you can apply the Mean Value Theorem?
I saw that you were working on another M. V. T. problem earlier. Have you reviewed the work done on that problem, so that you have an idea of what to do?
What is the derivative of y=e^(-7x)?
What are the values of y at the endpionts of the interval [0,9]?
Hope the answers to these questions will get you off to a good start in solving this homework problem.

Answer 2

yeah i have i tried it but i think the issue is that im not quit sure how to work with the e function so its messing my answere up

Answer 3

Have you looked up "exponential functions" on the Internet, or, more specifically, have you looked up "derivatives of exponential functions"? Or do you have a textbook to which to refer? (Gasp, some people still use textbooks!! :) )

The derivative of y=e^x is simply (dy/dx) = e^x (no, that's not a misprint).
What is the derivative of e^(2x)?

Answer 4

And what is the derivative of y=e^(-7x)? Hint: you must use the chain rule, because the exponent, -7x, is itself a function.

Answer 5

7e^(7x)? but when i plug it into the MVT what would be the next step?

Answer 6

The derivative of e^7x is 7e^(7x), but the function in question is not y=e^7x, but
y=e^(-7x). What's the derivative of that?

Answer 7

Do you have the M. V. T. equation in front of you? If not, please look it up.

Answer 8

-7e^(-7x) and yes i do

Answer 9

OK. the M. V. T. says, in effect, that the slope of the secant line connecting the two points determined by [0,9] is equal to the DERIVATIVE of the function at some value c. Our job is to find that value c, or values c1, c2, and so on.

Answer 10

Here f(x)=e^(-7x)
Please evaluate f(0) and f(9).
Then write the secant line slope:

f(9)-f(0)
-------
9 - 0

and set that expression equal to the derivative of y, evaluated at x=c.

this is exactly what the M. V. T. tells us.

Answer 11

f(0)= ?
f(9)= ?
f(9) - f(0) = ?

f(9)-f(0)
-------- = f'(c) = ?
9-0

Answer 12

You've already correctly calculated f'(x). Simply substitute c for x in f'(x).

Answer 13

Anna?

Answer 14

so im getting -1=-63e^(-7c)-e^63......where do i go from there?

Answer 15

Congrats! that looks good. Solving this equation for c is not an easy matter, however.

I'd write this equation as

Answer 16

\[\frac{ e ^{-63}-1 }{ -63 }=e ^{-7c}\]

Answer 17

and then I'd take the ln of both sides to eliminate e and to isolate (-7c).

Are you able to complete the work, that is, are you now able to solve for c?

Answer 18

does ln cancel the e?

Answer 19

Hint: It may be clearer if you re-write my equation, above, as\[\frac{ 1-e ^{-63} }{ 63 }=e ^{-7c}.\]

Answer 20

Yes, ln x and e^x are inverse functions, so one cancels the other.

Answer 21

-63 ln e - 1=-7c ln e

Answer 22

Anna, you have a fraction on the left side of my equation. Apply the log rule
ln a - ln b = ln (a/b) there.

Answer 23

I get:\[\ln(1-e ^{-63})-\ln (63) = -7c.\]

Answer 24

You can solve that for c, but the answer will not be a nice fraction. And you'll have round-off error.

But that's not your fault.

Lastly, you must check to determine whether or not your c value lies within the interval [0,9].

Answer 25

Anna, nice working with you. Hope this discussion has helped you.
I am very tired and need to hit the sack.
Do come back with this or other questions tomorrow, if you like.

Answer 26

ok thanks a lot for the help

Answer 27

:)